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Lecture Comments (22)

1 answer

Last reply by: Professor Hovasapian
Sat Jun 11, 2016 5:35 PM

Post by Suresh Kumar on June 11 at 10:30:43 AM


My biggest problem is that I'm confused on how to determine major species of the reaction.

Also, how are you able to determine which cations/anions/compounds don't really matter? In example 2, Cl- was crossed off, and I'm confused on why it was, although I know that you said that it's because it's part of HCl, which is a strong acid. Is this the only way to determine that it's not really that important?

I sort of understand why water is crossed off -- I recall that in previous lecture examples we used Ka to determine which ones didn't really matter. Since the Ka of water (Kw = 1.0 x 10^-14) is so small it was always crossed off. However, I don't really understand how we determine which one(s) to cross off now.

Thank you! The lectures are very helpful!

1 answer

Last reply by: Professor Hovasapian
Fri Mar 25, 2016 10:47 PM

Post by Nitin Prasad on March 17 at 07:22:12 PM

In example 1, if HF is formed, why don't you have to calculate its dissociation and formation of H+ from it?

1 answer

Last reply by: Professor Hovasapian
Sat Jan 10, 2015 7:27 PM

Post by Stephen Donovan on January 9, 2015

Isn't the shape of the aluminum hexahydrate ion (I think that's the right name, correct me if I'm wrong) due to d2sp3 hybridization on the aluminum?

0 answers

Post by Jianzhi Hu on June 23, 2014

Good, you fixed the mistake!


1 answer

Last reply by: Professor Hovasapian
Sat Apr 12, 2014 4:44 PM

Post by Tim Zhang on April 9, 2014

I encuntered a queation on disscociation of a weak acid in a solution. The Ph of this solution is larger than 7, which tells me this is a basic solution. I don't get it.....How come an acid can come up a basic solution.?

1 answer

Last reply by: Professor Hovasapian
Fri Mar 14, 2014 5:32 PM

Post by joebert binalinbing on February 23, 2014

when do we do the  approximation?.  example 1 (11:54)

1 answer

Last reply by: Professor Hovasapian
Tue Aug 13, 2013 2:54 AM

Post by Raymond Lanoria on August 13, 2013

How do we determine if it's a conjugate acid or base of a weak acid or base if something not familiar as F- is given?

4 answers

Last reply by: Professor Hovasapian
Sat Aug 3, 2013 3:05 PM

Post by Christian Fischer on August 2, 2013

Hi raffi, Great video!

Just one question: in example 2, the Ka of the acid reaction NH4(+) = NH3 + H(+) is 5.6*10^(-10) which is 3 magnitudes smaller than waters auto ionization. I know in previous videos you would discard this calculation because the concentration pf H+ produced is so small compared to 10^(-7). If this was a general problem and the goal was not to demonstrate how cations have an effect on ph, would my argument then be valid?

Have a great day

1 answer

Last reply by: Professor Hovasapian
Mon Oct 15, 2012 3:18 AM

Post by noha nasser on October 14, 2012

Hi prof. Raffi
i think there was a small mistake in the last example ( example 4 ).. the Ka of HSO4 i think its 1.2 X 10 to the -2 and not to the -6. you actually mentioned it in example 2 if i am not mistaken,With the ka equals to 1.2X10 to the -2 we get the Kb= 8.33X10 to the -13 :))

1 answer

Last reply by: Professor Hovasapian
Wed Oct 3, 2012 4:06 AM

Post by Saith Sanchez on September 30, 2012

14-5.58=8.41 still basic but wrong calculations I think

Related Articles:

Salts and Their Acid-Base Properties

  • Salts producing anions which are the conjugate bases of Weak acids (e.g. Nitrite Ion) react as weak bases, thus producing basic solutions.
  • Salts producing cations which are the conjugate acids of weak bases (e.g. Ammonium Ion) react as weak acids, thus producing acidic solutions.
  • If both are produced in the same solution, then compare the Kb of the anion with the Ka of the cation – whichever is bigger will dominate.

Salts and Their Acid-Base Properties

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Salts and Their Acid-Base Properties 0:11
    • Salts and Their Acid-Base Properties
    • Example 1
    • Example 2
    • Metal Ion and Acidic Solution
    • Example 3
    • NH₄F → NH₄⁺ + F⁻
    • Example 4

Transcription: Salts and Their Acid-Base Properties

Hello, and welcome back to, and welcome back to AP Chemistry.0000

Today, we are going to talk about salts and their acid-base properties.0004

Let's just jump on in, start with a definition, and start doing problems, because I think that is probably the best approach.0008

A salt is a generic term for an ionic compound.0015

Sodium phosphate; silver chloride; lead iodide; sodium chloride; you name it--potassium permanganate--these are all salts, because they are ionic compounds.0036

It is just a positive ion and a negative ion; we just call them all salts.0050

Salt, sodium chloride, is specific--just happens to take that name; but when we talk about salts, we are talking about any ionic compound.0054

Now, as you know (or as you should know), when you take an ionic compound and you drop it in water, it is either going to dissolve, or it's not.0062

If it dissolves, it breaks up into its free ions, and those ions are just floating around freely in the water; so let's write that down, actually.0069

When salts dissolve, their ions float around as free ions.0081

OK: so, for example, if we had something like sodium chloride (which is a solid, as you know--it's salt), if we drop it in water (an arrow with a little water on it--that means you have dissolved it in water), you end up creating Na+ + Cl-.0107

Now, I am not usually going to write the aq, the subscripts, but that is what that means--it just means "dissolved in water"--aq.0119

You have sodium ions floating around, and you have chloride ions floating around.0125

Well, often, when this happens, the anion of the salt that you drop into water and dissolve--it is actually going to be the conjugate base of a weak acid.0129

Let's say that again: Often (let's write it down), the anion--the negatively-charged ion, this one--will be the conjugate base of a weak acid.0141

Example: sodium fluoride--if we take sodium fluoride, and if we dissolve it in water, we produce sodium ion, and we produce fluoride ion; so there is F- floating around, and there is Na+ floating around.0166

Now, you have seen this F- before; it is the conjugate base of the weak acid hydrofluoric acid, which means--a conjugate base just means you have taken an acid and you have taken off the H.0182

The thing that you are left with--that is the conjugate base.0195

So, over here, I am going to write; you have seen it as (oops, let me do this in red to separate) this: HF is in equilibrium with H+ + F-.0197

There is a certain Ka associated with this; in fact, it's 7.2x10-4.0215

The Ka equals 7.2x10-4.0221

So notice, this time, it doesn't show up as the acid; you didn't take the acid and drop it in solution--you dropped a salt whose negative ion, the anion, happens to be the conjugate base of a weak acid.0226

When this happens--when the anion of a salt that you dissolve in water happens to be the conjugate base of a weak acid--it actually reacts with water as a base.0240

Because remember--notice--as an acid, this is a weak acid; a weak acid means the equilibrium is on the left; that means it wants to be this way--it doesn't want to be this way.0249

That means, if you put some free F- anywhere near some H+, or anywhere near a source of H+, it is going to take that H+, and it is going to move over in this direction until this equilibrium is established.0261

So, weak acid--strong conjugate base; a strong base means that it has a tendency, a very high affinity, for hydrogen ions--for protons--for hydrogen ions; it wants to take them.0275

When you take a salt and dissolve it in water, and all of a sudden create this free base, which happens to be the conjugate base of a weak acid (hydrofluoric acid), here is what it does: it actually behaves as a base now.0286

It does this reaction: it reacts with the water that is floating around in solution, and it actually takes the hydrogen ion from the water; it is acting as a base--it takes hydrogen ion.0301

This time, water is acting as the base; it becomes HF + OH-.0313

Notice what we have done: in this process, this base takes the H, creates hydrofluoric acid, and in the process it creates hydroxide ion.0321

When this happens, you create a basic solution; so if you took a neutral, normal water, which is pH 7, and if you drop in some sodium fluoride, well, the fluoride ion will pull hydrogens off of the water, creating hydroxide ion.0330

The pH of that solution is going to go up; it is going to become a basic solution--that is what a basic solution is: it's when the concentration of hydroxide ion is higher than usual.0345

So, any time you have a salt where the anion happens to be the conjugate base of a strong acid...conjugate base of a weak acid; forgive me--where it happens to be the conjugate base of a weak acid, it is going to react as a base in this reaction.0356

People call this a hydrolysis reaction; I don't really like that name--just know that now, this conjugate base is going to act as a base.0374

You have seen this reaction before; this is the reaction of a base with water.0382

There is a Kb associated with this.0387

There is a Kb associated with this: it is HF, OH-, over F- (because water is liquid water).0390

Well, how do we know what the Kb is?0403

Here is the best part about it: the Kb, the relationship--we know the Ka already, but what is the relationship between the Ka and the Kb, since now the F- is actually behaving as a base, and not this way, as part of the weak acid equilibrium?0407

Now, it is involved in a basic equilibrium, where it is actually pulling off a hydrogen ion.0422

Here is the relationship--all Kas and Kbs are related by this equation: Ka, times the Kb, is equal to Kw.0429

So in this case, if you want to do a problem with this, and you need to treat this as a Kb problem, base problem, because this reaction is that of a base with water, all you do is: you take the Kb is equal to Kw over Ka for that species.0438

This is 10-14; in this particular case, it's 7.2x10-4, and you find the Kb, and you run this problem as a base problem.0459

We have already done it: weak acids, weak bases--that is all that is going on here.0470

So, let's actually do a problem, and it will make sense.0474

Let's see...OK; so, Example 1: Calculate the pH of a 0.35 Molar sodium fluoride solution.0479

The Ka is equal to (sorry, I had better say which Ka of what)...the Ka of hydrofluoric acid (the actual acid, the weak acid, where the base is coming from) is 7.2x10-4.0507

OK, what do we always do first?--we check the major species in solution to decide what chemistry is going to dominate.0529

Major species: well, you have sodium fluoride: sodium fluoride is completely soluble--remember the solubility chart from earlier in the year?0535

If you haven't memorized it, not a problem--just check it out; sodium fluoride, alkali metal, halogen--completely soluble.0545

That means what you have floating around in solution is sodium ion; you have fluoride ion; and you have H2O.0552

Well, we notice sodium ion doesn't do anything; it doesn't affect anything.0560

However, the anion happens to be the conjugate base of a strong acid, HF.0565

I keep saying "strong acid"; what is wrong with me?--weak acid, weak acid, weak acid.0571

So, F- is the conjugate base of a weak acid, hydrofluoric acid.0577

Well, it's the conjugate base of a weak acid; so what it is going to do--it is actually going to react as a base with the water, the following reaction.0582

It is going to be: F- + HOH goes to HF + OH-.0591

That is the reaction; this reaction is the reaction of a base with water.0608

There is a Kb associated with it.0614

We want to find the pH of this; well, now that we have our reaction, that we know what reaction is going to take place, we do our ICE chart.0616

Well, what is the initial concentration of the F-?--well, since we have full dissociation, it's 0.35.0626

Water doesn't matter; there is no HF formed yet--this is before anything happens, and there is no OH- before anything happens.0634

A certain amount of F- is going to disappear; as a species, it's going to react with H to become HF.0642

It is going to disappear; water doesn't matter; that means HF is going to show up, and OH- is going to show up.0649

There is absolutely nothing under the sun when it comes to these; we have done these several times--we know how it works, but now, because this is a base reaction, we need the Kb.0659

Kb is equal to Kw over Ka, equals 1.0x10-14, divided by 7.2x10-4, and we get a Kb of 1.4x10-11.0670

So now, we do 1.4x10-11 is equal to x, times x, divided by 0.35-x.0696

Well, look how small this is; you know what, x is going to be pretty small, so chances are, we can do the approximation.0709

It equals x2 over 0.35.0716

When we solve for x, we get x=2.2x10-6; but notice, this was not hydrogen ion concentration; we created a base in this reaction: x is equal to the OH- concentration, which implies that the pOH is equal to the negative log of this.0722

The pOH ends up being 3.2 (was that correct?--yes), and (wait, is that...yes, it is) then the pH is equal to 14 minus the pOH, which equals 10.8.0742

Sure enough, a pH of 10.8 means that it is a basic solution.0773

We had a salt; the anion was the conjugate base of a weak acid; therefore, it is going to react as a base with the water in a standard base reaction.0778

That is what bases do: bases take hydrogen from water to create hydroxide ion.0794

The equilibrium expression for that: we said it's a Kb, not a Ka; it's not an acid dissociation--it is a base association, if you will; it's a base constant.0799

Well, we have the Ka; normally, hydrofluoric acid is listed as a Ka, because it behaves mostly as an acid.0810

Therefore, the Ka is listed; but the relationship between Ka and Kb is Ka times Kb equals 10-14, which is Kw.0818

We solve for the Kb; we treat it like any other weak base problems; and we solve it; we get the pOH, in this case, because it's a base--because we want the pH, we take 14 minus that, 10.8.0826

I hope that makes sense.0839

OK, let's do Example 2: oh, actually, before that, let me let's stop there, and now let me go back to blue ink...0841

Now, if you have a salt (now we are going to talk about the cation; we mentioned the anion; now we are going to talk about the cation) where the cation is the conjugate acid of a weak base, then this cation will act as an acid and create hydrogen ion--create an acidic solution.0858

So, you have to watch the salt; you take a look at the salt, dissolve it; you take a look now--you not only look at the anion--you look at the cation as well.0919

For the first one, we said if the anion happens to be the conjugate base of a weak acid; now, if the cation happens to be the conjugate acid of a weak base.0928

It is going to act as an acid, and it is going to produce an acidic solution.0944

Let's do an example, and I think it will make sense.0949

And again, it is the chemistry that you want to understand: you want to take a look at the species and decide how it is acting.0954

That is really what is going on; in this previous problem, we saw that, when we dissolved the sodium fluoride, we have F- floating around freely.0962

Well, what is F- going to do?--F- happens to be the conjugate base of a weak acid, so it is going to actually behave as a base; it is going start taking Hs away from water.0969

If you write down the reaction, everything should fall out.0980

OK, calculate the pH of a 0.10 Molar ammonium chloride solution; the Kb of NH3 equals 1.8x10-5.0985

OK, our major species: well, we know that anything that involves ammonium and chloride is going to be fully soluble; therefore, what is floating around in solution is ammonium ion, chloride ion, and H2O.1013

Chloride--let's look at the anion first--chloride is the conjugate base of a strong acid, HCl.1031

So, Cl is not going to take any H from anything; it is just going to float around freely--we can ignore it.1040

However, NH4+, ammonium, is the conjugate acid of a weak base.1047

How do we find the conjugate acid?--just stick a hydrogen ion onto it...of the weak base ammonia.1056

Ammonia is the weak base; its conjugate acid is that, right?--because it comes from ammonia.1062

When you put ammonia in water, it takes a hydrogen from water; it becomes ammonium, and it creates hydroxide.1074

However, in this case, we didn't just drop ammonia into solution; we dropped the actual ammonium ion into solution as a salt.1084

We dropped it as a salt; now, there is just free ammonium ion floating around--all ammonium ion.1093

Well, ammonium ion is the conjugate acid of a weak base, which is ammonia.1099

Therefore, it is going to now behave as an acid.1104

Its equilibrium is going to be this: it is going to actually dissociate into H+ +, plus NH3.1107

There is a Ka associated with this, because this is an acid dissociation reaction: it is something that has a hydrogen ion to give up.1122

It gives it up, and it creates its other side; so now, this Ka is equal to Kw over the Kb.1130

Kb is the Kb that we have for ammonia, the conjugate base of that.1138

So now, let's solve the problem.1144

NH4+ is the dominant species; it is the conjugate acid of a weak base--it is going to behave as an acid, so we write it: behaving as an acid.1147

We are going to create H+; we are going to create an acidic solution, because we dropped it in water that was neutral.1164

We have Initial, we have Change, and we have Equilibrium; well, the initial concentration of NH4 is 0.10.1171

There is no H to start with; there is no ammonia to start with; this is -x; this is x; this is x; +x; this is 0.10-x; this is +x; this is +x.1180

Now, it's behaving as an acid, so we need a Ka.1194

Well, a Ka is equal to Kw over Kb.1199

That is equal to 1.0x10-14, divided by 1.8x10-5; the Ka for this reaction is equal to 5.6x10-10.1204

That tells me, this's a small number; it is a weak acid.1222

But, it is still an acid--it is going to produce some's weak, but it still behaving as a weak acid--it's going to give up its H+ into floating around freely in the water; that is the acidic solution.1228

The pH of the solution is going to drop.1239

So now, let's take 5.6x10-10 is equal to x, times x, over 0.10-x, which is approximately equal to x squared over 0.10; we end up with x=7.4x10-6, which does equal the hydrogen ion concentration, because here, we are directly forming hydrogen ion, and therefore, the pH of this is the negative log of that, which ends up being 5.1.1241

Sure enough, 5.1 is an acidic solution.1279

So there you have it: recap: if you have a salt floating around in solution (if you have a salt and it dissolves), if the anion is the conjugate base of a weak acid, it will create a basic solution.1284

You treat it as a base equilibrium that will react with water, in other words--pull off the hydrogen ion.1301

If the cation is the conjugate acid of a weak base, it will behave as an acid--you write the acid equilibrium and use the Ka to solve it as a weak acid problem, like you have done before.1307

It will create an acidic solution.1318

A second type of cation--a second type of species (you know, it might be nice if I actually wrote my words properly here--what do you think?) that creates an acidic solution--is a highly-charged metal ion.1321

A good example is aluminum 3+; 3+ is a pretty high charge; you are also going to find solutions like, for example, chromium 6+, manganese 5+, 4+...things that have a really high charge after about 1 or 2.1362

Here is what is going on: if you take a salt like aluminum chloride, I'm going to actually write out the chemistry of what happens, because I want you to see what happens--how the species forms in solution, and then how we treat that species in solution, just like any other weak acid.1376

If you take aluminum chloride, and you dissolve it in water--well, you know that aluminum chloride is completely soluble, so you are going to end up with aluminum 3+, plus 3 chloride ions--it completely dissociates into its free ions.1395

Chloride doesn't do anything; it's the conjugate base of a strong acid, so it doesn't behave as a base--it just floats around.1408

However, aluminum (because it is so highly charged, and because it is in water)--the water molecules, the actual molecules themselves, surround the aluminum; and in the case of aluminum, you get something that looks like this.1415

I am actually going to draw the structure, so you see it; I'm going to write it out first.1431

Aluminum (so I'm going to bring it over here)--the aluminum 3+ actually associates with 6 water molecules (and I'm going to draw the lone pairs on them) to create this species called a complex ion.1435

Al(H2O)6, and the whole charge on the species is 3+; this is usually how we write it.1456

Any time you have a metal that is surrounded by, in this case, water (we will just deal with water for the moment), and this actually looks like something--you have an aluminum ion, and you have an H2O, an H2O...I'm going to draw this 3-dimensional structure.1465

I really shouldn't, because you really don't have to know this for the time being; we may talk about it a little bit later, towards the end of the course, but I think it's important; it's nice--there is nothing here that you shouldn't be able to sort of visualize.1487

There are these 6 water molecules: OH2, OH2, OH2, OH2...and it is actually the oxygen, believe it or not, that is attached to the aluminum here.1499

In a normal bond, it is just surrounding it; so you have these 6 water molecules, and they are arranged; four of them are arranged in a plane; so aluminum is in the middle; one here, here; one is here; and back here.1512

These dashed lines mean it is going back; it's facing away from you.1525

These wedges mean it's coming towards you; these straight lines mean there is one on top and one on bottom.1528

What you have is the following: what you have is aluminum, in the center; you have a water molecule here, a water molecule here, a water molecule back here, a water molecule back here; one up here, and one down here.1535

There are six of them around it; and this whole species is carrying a 3+ charge; well, of course it's carrying a 3+ charge--aluminum is 3+, and water is a neutral molecule, so when they aggregate (when they sort of grab onto aluminum, if you will--6 of them), you create this thing called a complex ion.1548

Well, here is what happens with this complex ion: now I'm going to take this complex ion; as it turns out, it's so highly charged that it actually ends up pulling a lot of the electrons towards itself, and actually creates an acidic hydrogen.1569

One of these hydrogens, believe it or not, actually comes off.1584

And here is the chemistry of it (and this is what is important: the structure you don't have to understand; the chemistry is what is going on): in solution, this is produced.1588

When you drop aluminum chloride into water, aluminum ion is formed; water is attracted to the aluminum ion, and six water molecules arrange themselves in a pattern around the aluminum to create this complex ion.1598

This complex ion has an acidic hydrogen; it gives it up.1612

Here is how it gives it up: Al(H2O)63+ (I'm going to leave off the brackets; I think it's just extra symbolism that is not necessary)--it dissociates into H+ + Al; now there is just a hydroxide attached to it, and there is 5 waters.1615

One of the waters has lost its hydrogen; now it's 2+.1639

Notice, this is just a standard dissociation of an acid.1644

You have this species that has this H that it can give up; it gives it up; it's right there.1649

The rest of it--it doesn't even matter what it is; all that matters is this.1656

You have a species that has a hydrogen ion; it gives up that hydrogen ion to create some conjugate base; this is what is important.1660

You treat this like any other weak acid; there is a Ka associated with this--we measured it, and the Ka of this happens to be 1.4x10-5.1667

That is it; this is just HA dissociating into H+ + A-; this A---yes, it happens to be a very, very complex-looking thing (we call it a complex ion), but you treat it the exact same way.1679

Don't let the makeup of the thing that you are discussing confuse you; it's the chemistry that matters--the chemistry is just: some species gives up a hydrogen ion, and then ends up as something else.1694

The hydrogen ion is what is important; the equilibrium is handled the exact same way.1708

So now, let's do a problem.1713

Example 3: This is what is important in science--you need to understand what is going on underneath.1717

The individual identities of the species--they don't really matter, as far as what is going on; they matter for the individual case that you happen to be dealing with, as a researcher, as a lab scientist, as a doctor, whatever it is, but the chemistry is all the same.1726

It is still just some species, some acid, that has a hydrogen to give up, and it gives it up.1742

The mathematics is handled exactly the same way; the species, the identity, is entirely irrelevant.1747

It is entirely irrelevant; this is what we want you to do.1752

Our ideal is to get you to think abstractly, to think big-picture; if you can handle the big picture, you will know what is going on with the little picture; the little details are just incidental--they change from problem to problem.1756

But, the big picture doesn't change; that is what is important.1769

In science, what you want to concentrate on is what doesn't change.1772

The things that change...well, they are incidental.1777

That is when you know something is important--if something is not changing, that is what you want to concentrate on.1780

It is the chemistry that is important; OK.1785

Sorry about that lecture.1788

So, Example 3--we have: Calculate the pH of a 0.010 Molar AlCl3 solution.1790

OK, so the first thing we do: major species. 1808

Well, here is what we know: when we have some aluminum salt that is dropped in water, the aluminum is going to float around freely as ion; that is the first thing that is going to happen--the aluminum chloride is going to dissociate.1813

Aluminum is...the water molecules are going to aggregate around aluminum, 6 of them are, and they are going to form the species Al(H2O)63+.1827

Now, it isn't important that you know the name of it, but this is called hexaaqua-aluminum (3).1841

You will actually do the naming towards the end of the course, when you talk about coordination compounds, but this is the species that actually forms in solution.1846

If we took a picture of the solution, that is the species that we find.1854

We find every aluminum ion surrounded by six water molecules, and that whole thing is carrying a 3+ charge.1858

The only other species in there is water.1864

Well, it is true--water does contribute some hydrogen ion--but it is Ka of 10-14; this one has a Ka of 1.4x10-5.1867

10 to the negative 5 is a lot bigger than 10 to the negative 14, so water can be ignored.1881

This is the dominant species in the water that will control the pH of the solution.1885

We know what this does: it behaves as an acid.1893

Al(H2O)63+ dissociates into H+ + AlOH(H2O)52+--just an acid dissociation; that is it.1896

This is being created; let's do Initial; let's do Change; let's do Equilibrium.1919

What is the initial concentration?--well, all of the aluminum is dissolved; that means all of this is formed; 0.010--there is nothing formed yet; there is nothing formed yet.1924

A certain amount is going to dissociate--that is how much is going to show up of the other species.1935

0.010-x, +x, +x; we have the Ka; we have the equilibrium expression; so we just put it in.1940

1.4x10-5 (I hope you're not getting sick of these problems; I know it's just over and over--it's the same thing; that is nice--we like patterns) equals x times x, divided by 0.010-x.1951

We can approximate this with x squared over 0.010; now, you might think to yourself, "Well, wait a minute; 0.010 is pretty small, and x...we are talking about 10 to the negative 5 here...maybe."1972

As it turns out, when you check the validity, the error ends up being about 3.7%; we are still below 5, so we are good; this is a perfectly good approximation.1987

So, x is equal to 3.7x10-4, which equals the hydrogen ion concentration; that implies that the pH is equal to 3.43.1998

3.43--OK, I think I want to write this a little bit slower, so that all these wacky lines don't show up--equals 3.43.2013

How is that?2026

So notice, it's handled exactly the same way; the identity of the species is irrelevant--it's behaving as an acid; it's giving up a hydrogen ion.2027

It's a weak acid, 1.4x10-5; there is an equilibrium; we have to use an ICE chart.2036

We are done; that is nice; OK.2043

Now, our final little situation here: what if we have this situation?2048

What if we have the situation: NH4F--what if we have ammonium fluoride?--it is a perfectly valid salt.2052

Ammonium chloride...we can form ammonium fluoride.2070

Salt--you drop it into water; what happens?--well, it's going to dissolve, if it's completely soluble; it's going to dissolve into NH4+ and F-.2074

Well now, I have a little bit of a problem: we have an anion which is the conjugate base of a weak acid, hydrofluoric acid; so it is going to behave as a base, and it is going to pull hydrogen off of water to produce hydroxide ion.2083

So, it is going to create a basic solution; but, we have ammonium ion also floating around in solution.2099

It is the conjugate acid of a weak base, ammonia, and it is going to behave as an acid, as a weak acid, itself.2106

It is going to give up its hydrogen ion to create an acidic solution.2115

The F- is going to go ahead and create basic solution; this is going to create an acidic solution; well, what is the final solution going to be--acidic or basic? How do we decide?2118

Well, when you have a situation where both of the ions are species that react, and one produces base; one produces water, it gets very, very complicated--that is the short answer.2128

The equilibrium gets complicated, and you actually will be dealing with stuff like this, if you go on to study analytical chemistry.2141

If you are a chemistry major, usually in your third year, you will take an analytical chemistry course, and there are ways to handle this mathematically.2147

Pretty complex; for our purposes, we just want to be able to sort of give a qualitative answer.2154

We want to be able to say is the solution acidic or basic, without specifying what the pH is.2159

That is actually very, very easy to do.2164

It comes down to this: you do a quick test--if (I'm actually going to write this a little further to the left--excuse me) the Ka for the acidic ion (meaning this one) is bigger than the Kb for the basic ion, which is this one, well, the solution is acidic.2168

In other words, if the Ka for this is bigger than this, that means that the equilibrium for this is farther to the right; it produces more hydrogen ion than this produces hydroxide ion.2207

Therefore, there will be more hydrogen ion in solution; therefore, the solution will be acidic.2219

That is what this is saying: so, you need to find the Ka of this; you find the Kb of this; remember, Kb is 10 to the 14 over Ka of the acid, and this Kb is 10 to the 14 over Ka.2223

I'm sorry; Ka is 10-14/Kb for the conjugate base, what we did earlier.2238

You compare the two; the bigger one--that will dominate.2245

If this is bigger, it will be acidic; if this is bigger, it will be basic.2249

So, if the Ka is less than the Kb (the Ka for the acidic ion, less than the Kb for the basic ion), then your solution is going to be basic.2254

And of course, the last possibility (always three possibilities when it comes to ordering: less than, greater than, or equal to): If the Ka is equal to the Kb, well, you know exactly what that is; that is going to get a neutral solution.2265

And now, let's do our final example: Example 4: Will a solution of aluminum sulfate (AlSO4, Al2(SO4)3) be acidic or basic?2283

Well, Al2(SO4)3 dissociates into Al2, aluminum 3+, plus 3 SO42-, so yes; we have an anion--negative ion--which is the conjugate base of a weak acid.2315

The weak acid, in this case--just add one H.2339

OK, it's HSO4-; not H2SO4; it's HSO4---add one H.2343

And aluminum happens to be that thing that forms that species, Al, the hexaaqua-aluminum (3); so this is going to create a basic solution; this is going to create an acidic solution; we need to compare the two.2350

So, we want to know the Ka of this; the Ka--we already know that one; that is going to be 1.4x10-5.2369

And we want to know the Kb of this; the Kb of this is 10 to the -14, over the Ka of this, which is 1.2x10-6.2383

I hope you guys saw what I did; I found out the species that are floating in solution; this--the conjugate acid of that is the HSO4.2401

This actually just forms this species; we have the Ka of this; we find the Kb; we don't take the H2SO4; we add one H to it, OK?--one H at a time.2411

It dissociates one at a time; it associates one at a time.2422

We get a Kb, is equal to 1.3x10-13; well, this is hugely bigger than this, which means that our solution will be acidic.2427

This aluminum, this aluminum species, will dominate the acidity of the solution; we will get an acidic solution; and that is how you handle it.2444

So, thank you for joining us here at to discuss the acid-base properties of salts.2454

In our next lesson, I'm going to close off with just a brief discussion of some oxides, and then we will go ahead and move on to further aspects of acid-base equilibria.2459

We will talk about buffer solutions and titration curves.2470

Take care; see you next time; goodbye.2472